Please solve q20!

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Let O be the centre of a circle and AB, CD be the two chords. Let PQ be the diameter bisecting chord AB and CD at L and M respectively.
 

L is the mid- point of AB

So   OL ⊥ AB  ⇒  ∠ALO = 90 ̊  

Similarly,   ∠CMO = 90 ̊

∠ALO = ∠CMO =  90 ̊

... co-interior angles are supplementary

So  AB ∥ CD.

Hence proved.

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